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The matrix logarithm near the identity maps $Sp(n)=\{A\in U(2n):AJ{A}^t=J\}$ onto an open set of its real compact symplectic Lie algebra $\{X:X^{\ast }=-X,XJ+J{X}^t=0\}$. Its inverse is the matrix exponential. These statements follow by applying the logarithm to $A^{\ast }=A^{-1}$ and $A^t=J^{-1}{A}^{-1}J$, and differentiating $e^{tX}J{e}^{t{X}^t}$. Left translations provide manifold charts at every group element. In block form $X=\left(\begin{array}{cc}P& Q\\ -\bar{Q}& \bar{P}\end{array}\right)$ with $P^{\ast }=-P$ and $Q^t=Q$, giving real dimension $n^2+n(n+1)=n(2n+1)$.

A Lie group is a finite-dimensional smooth manifold with a group structure for which multiplication and inversion are smooth. The group defined here is the compact symplectic group $Sp(n)=U(2n)\cap Sp(2n,\mathbb{C})$, rather than the full complex symplectic group. Also, the displayed matrix expression requires $B$ to be $2n\times 2n$; the printed $n\times n$ size is incompatible with $J$.

Since $J^2=-I$, we have $J^{-1}=-J$. The matrix exponential commutes with conjugation, as follows term by term from its absolutely convergent power series. ConsequentlyTo construct logarithm charts for the compact symplectic group, consider the real vector spaceDifferentiating the defining identities at $I$ gives precisely these conditions. Conversely if $X\in \mathfrak{k}$, $e^X$ is unitary andso $e^X\in Sp(n)$. These are the infinitesimal conditions of the compact symplectic Lie algebra.

Near $I$, the convergent matrix logarithm series $\log (I+C)={\sum }_{m\ge 1}(-1{)}^{m+1}{C}^m/m$ is smooth and inverse to the matrix exponential near zero. These local inverses respect transpose, conjugate transpose, and conjugation; also $\log (A^{-1})=-\log A$ when both matrices are sufficiently close to $I$. Shrink their neighborhoods accordingly. If $A\in Sp(n)$ there, unitarity givesThe symplectic identity is equivalent to $A^t=J^{-1}{A}^{-1}J$, so, putting $X=\log A$, it gives $X^t=-J^{-1}XJ=JXJ$, equivalently $XJ+J{X}^t=0$. Thus this local logarithm restricts to a bijection between a neighborhood of $I$ in $Sp(n)$ and an open neighborhood of zero in $\mathfrak{k}$. Its inverse is the restricted matrix exponential. These restrictions are manifold charts; left multiplication translates them to every $A_0\in Sp(n)$ via $A\mapsto \log (A_0^{-1}A)$. The chart overlaps are smooth compositions of multiplication, exponential and logarithm. The subspace topology is Hausdorff and second countable, inherited from the finite-dimensional matrix space, so these charts give a smooth manifold.

Closure under products and inverses follows from $AJ{A}^t=J$ and unitarity. Matrix multiplication is polynomial in real and imaginary entries, and inversion on the unitary group is $A\mapsto A^{\ast }$, a real linear operation. Their restrictions are smooth in the charts just constructed. Hence $Sp(n)$ is a Lie group without needing a closed-subgroup theorem.

Write $X$ in $n\times n$ blocks. The two infinitesimal conditions giveThe skew-Hermitian matrix $P$ has $n^2$ real parameters; the complex symmetric matrix $Q$ has $n(n+1)/2$ complex parameters, hence $n(n+1)$ real parameters. ThereforeFor $n=1$, any two-by-two matrix satisfies $AJ{A}^t=(\det A)J$, so the group is $SU(2)$. Explicitly,The rows are orthonormal and the determinant is one; conversely unitarity and determinant one force this form. This is the SU(2) as the three-sphere parametrization. The map and its inverse, extraction of the first row, are smooth. Thus $Sp(1)$ is diffeomorphic to $S^3$.

Over $\mathbb{R}$ or $\mathbb{C}$, a Lie algebra is a vector space with a bilinear map $[,]$ which is alternating and obeys the Jacobi identity:Bilinearity and alternation imply $[X,Y]=-[Y,X]$.

For a Matrix Lie group, identify the tangent space $\mathfrak{g}=T_{I}G$ with derivatives ${\gamma }^{\prime }(0)$ of smooth curves through $I$. Product curves show that the sum of two such derivatives is again tangent, and reparametrization supplies scalar multiples. In particular this is a real vector space, even when the matrices have complex entries. For $X,Y\in \mathfrak{g}$, choose a curve $\gamma (s)$ with ${\gamma }^{\prime }(0)=X$. Conjugating a curve with derivative $Y$ shows thatDifferentiate this curve in the finite-dimensional vector space $\mathfrak{g}$. The result isThe matrix commutator is bilinear and alternating, and expanding the six terms proves its Jacobi identity. Thus this construction gives the Lie algebra of a matrix Lie group, with the appropriate bracket, using actual group curves rather than an assumed commutator closure.

For the unitary group, differentiating $\gamma (t{)}^{†}\gamma (t)=I$ gives $X^{†}+X=0$. Conversely, if $X^{†}=-X$, then $e^{tX}$ is unitary and is a curve with derivative $X$. ConsequentlyThis is the unitary Lie algebra. The diagonal entries are purely imaginary, contributing $N$ real parameters, and each upper off-diagonal entry contributes two real parameters. A matrix-unit basis of the unitary Lie algebra isThese $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$ anti-Hermitian matrix units and their combinations are linearly independent over $\mathbb{R}$ and span every allowed entry.

The symplectic stabilizer is a subgroup: the identity preserves $J$, and if $MJ{M}^T=J$ and $NJ{N}^T=J$, thenMultiplying $MJ{M}^T=J$ on the left by $M^{-1}$ and on the right by $M^{-T}$ gives $M^{-1}J{M}^{-T}=J$. Products and inverses remain unitary. This is the compact symplectic group, often denoted $USp(2n)$ or $Sp(n)$.

Differentiating the stabilizer equation gives $XJ+J{X}^T=0$. For $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$, this saysTogether with $X^{†}=-X$, these are equivalentlyThese conditions are also sufficient: $e^{tX}$ is unitary, andHence it stays in the subgroup. They characterize its compact symplectic Lie algebra without adding any trace condition. In fact the trace automatically vanishes. The free anti-Hermitian block $A$ has $n^2$ real parameters, and the complex symmetric matrix $B$ has $n(n+1)$ real parameters. Thus

Here is a complete matrix-unit basis of the compact symplectic Lie algebra. The generators supplying $A$ areThe two families supplying the real and imaginary parts of $B$ are, for $1\le i\le j\le n$,The denominator merely avoids double-counting diagonal entries. Each displayed generator obeys both defining tangent conditions. The $H,P,Q$ generators form a real basis of the allowed $A$ blocks, and the $U,V$ generators form a real basis of the complex symmetric $B$ blocks. Thus they are independent and their total number is $n(2n+1)$. They give all generators required by the real compact algebra, including the case $n=1$.